Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<cstdio>
using namespace std;
double dp[3000][3000]={},p,ans=0.0;
int main(){
    
    int n;
    scanf("%d",&n);
    dp[0][0]=1.0;
    for(int i=0;i<n;i++){
        scanf("%lf",&p);
        dp[i+1][0]=dp[i][0]*(1-p);
        for(int j=1;j<=n;j++){
            dp[i+1][j]=dp[i][j]*(1-p)+dp[i][j-1]*p;
        }
    }
    for(int i=n/2+1;i<=n;i++){
        ans+=dp[n][i];
    }
    printf("%.10f\n",ans);
}